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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2017 · Mathematics · Quadratic Equation

If α and β are the roots of the equation x^2+b x+c=0 with b ≠ 0 ,...

If cot⁡α\cot \alpha and cot⁡β\cot \beta are the roots of the equation x2+bx+c=0x^2+b x+c=0 with b≠0b \neq 0, then the value of cot⁡(α+β)\cot (\alpha+\beta) is

Options

  1. Ac−1b\frac{\mathrm{c}-1}{\mathrm{ b}}
  2. B1−cb\frac{\mathrm{1}-c}{\mathrm{ b}}
  3. Cbc−1\frac{\mathrm{b}}{\mathrm{c}-1}
  4. Db1−c\frac{\mathrm{b}}{\mathrm{1}-c}

Correct answer

B. 1−cb\frac{\mathrm{1}-c}{\mathrm{ b}}

Step-by-step solution

 Given equation, x2+bx+c=0Roots are cot⁡α,cot⁡βSum of roots =cot⁡α+cot⁡β=−bProduct of roots =cot⁡α⋅cot⁡β=ccot⁡(α+β)=cot⁡α⋅cot⁡β−1cot⁡β+cot⁡α=c−1−b=1−cb\begin{aligned} & \text{ Given equation, } \mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0 \\ & \text{Roots are } \cot \alpha, \cot \beta \\ & \text{Sum of roots }=\cot \alpha+\cot \beta=-b \\ & \text{Product of roots }=\cot \alpha \cdot \cot \beta=c \\ & \cot (\alpha+\beta)=\frac{\cot \alpha \cdot \cot \beta-1}{\cot \beta+\cot \alpha}=\frac{c-1}{-b}=\frac{1-c}{b}\end{aligned}
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